Question

Breaks are applied to the car to produce negative acleration 6m/s sq . if the car stops after 2 seconds. find the initial velocity.
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Answer 1

Answer:

v=0

a=-6

t=2s

u=?

v=u+at

u=v-at

u=0-(-6)×2

u=12m/s

Answer 2

Answer:

V=0 because after applying brakes the car comes to rest.

t=2 s

a=-6ms⁻²

u=?

So, we have to find the initial velocity of the car.

For this we have to use the first equation of motion. We have;

v=u+at

0=u+(-6)x2

0=u-12

u=12 m/s

Now to find the distance we can use either the second or third equation of motion.

I will show you in both ways

By using the second equation, we have:

s=ut+1/2.at^2

24+1/2.(-6)x4

24-12=12m

By using the third equation, we have:

v²-u²=2as

0-144=2x(-6)xs

-144=-12s

12s=144

s=12m